Problem: Find $\lim_{x\to -4}\dfrac{7x+28}{x^2+x-12}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $7$ (Choice C) C $-1$ (Choice D) D The limit doesn't exist
Answer: Substituting $x=-4$ into $\dfrac{7x+28}{x^2+x-12}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{7x+28}{x^2+x-12}$ can be simplified as $\dfrac{7}{x-3}$, for $x\neq -4$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-4$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{7x+28}{x^2+x-12}=\dfrac{7}{x-3}$ for all $x$ -values in the interval $(-5,-3)$ except for $x=-4$. Therefore, $\lim_{x\to -4}\dfrac{7x+28}{x^2+x-12}=\lim_{x\to -4}\dfrac{7}{x-3}=-1$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -4}\dfrac{7x+28}{x^2+x-12}=-1$.